| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward hypergeometric distribution problem requiring basic combinatorics to construct the probability table (choosing 0, 1, or 2 cats from 7 animals), then applying the standard variance formula Var(X) = E(X²) - [E(X)]². The calculations are routine with small numbers and no conceptual challenges beyond recognizing the setup. |
| Spec | 2.04a Discrete probability distributions5.01a Permutations and combinations: evaluate probabilities5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| \(X\) | 0 | 1 |
| Prob | \(\frac{2}{7}\) | \(\frac{4}{7}\) |
| B1 | Prob distribution table drawn, top row correct with at least one probability \(0 < p < 1\) entered | |
| \(P(0) = \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} = \frac{2}{7}\) (0.2857) | B1 | One probability correct (need not be in table) |
| \(P(1) = \frac{2}{7} \times \frac{5}{6} \times \frac{4}{5} \times {}^3C_1 = \frac{4}{7}\) (0.5713) | B1 | Another probability correct (need not be in table) |
| \(P(2) = \frac{2}{7} \times \frac{1}{6} \times \frac{5}{5} \times {}^3C_2 = \frac{1}{7}\) (0.1429) | B1 | Values in table, all probs correct (to 3SF) or 3 probabilities summing to 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Var}(X) = 1 \times \frac{4}{7} + 4 \times \frac{1}{7} - \left(\frac{6}{7}\right)^2\) | M1 | Unsimplified correct numerical expression for variance or *their* probabilities from (i) \(0 < p < 1\) in unsimplified variance expression |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{20}{49}\) or 0.408 | A1 | Correct answer (0.40816…) nfww; final answer does not imply the method mark |
## Question 4(i):
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| Prob | $\frac{2}{7}$ | $\frac{4}{7}$ | $\frac{1}{7}$ |
| B1 | Prob distribution table drawn, top row correct with at least one probability $0 < p < 1$ entered
$P(0) = \frac{5}{7} \times \frac{4}{6} \times \frac{3}{5} = \frac{2}{7}$ (0.2857) | B1 | One probability correct (need not be in table)
$P(1) = \frac{2}{7} \times \frac{5}{6} \times \frac{4}{5} \times {}^3C_1 = \frac{4}{7}$ (0.5713) | B1 | Another probability correct (need not be in table)
$P(2) = \frac{2}{7} \times \frac{1}{6} \times \frac{5}{5} \times {}^3C_2 = \frac{1}{7}$ (0.1429) | B1 | Values in table, all probs correct (to 3SF) or 3 probabilities summing to 1
**Total: 4 marks**
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## Question 4(ii):
$\text{Var}(X) = 1 \times \frac{4}{7} + 4 \times \frac{1}{7} - \left(\frac{6}{7}\right)^2$ | M1 | Unsimplified correct numerical expression for variance or *their* probabilities from (i) $0 < p < 1$ in **unsimplified** variance expression
$= \frac{8}{7} - \left(\frac{6}{7}\right)^2$
$= \frac{20}{49}$ or 0.408 | A1 | Correct answer (0.40816…) nfww; final answer does **not** imply the method mark
**Total: 2 marks**4 Mrs Rupal chooses 3 animals at random from 5 dogs and 2 cats. The random variable $X$ is the number of cats chosen.\\
(i) Draw up the probability distribution table for $X$.\\
(ii) You are given that $\mathrm { E } ( X ) = \frac { 6 } { 7 }$. Find the value of $\operatorname { Var } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2018 Q4 [6]}}